# Answer to Question #10632 in Mechanics | Relativity for Megha reddy

Question #10632

A farmer moves along the boundary of a square field of side 10m in 40 seconds. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Expert's answer

As a farmer moves along the boundary of a square field of side 10m in 40 seconds, then his velocity is

V = 10/40 = 0.25 m/s.

So, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds (or 60·2+20 = 140 seconds) from his initial position will be

S = 140·0.25 = 35 m.

V = 10/40 = 0.25 m/s.

So, the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds (or 60·2+20 = 140 seconds) from his initial position will be

S = 140·0.25 = 35 m.

## Comments

Assignment Expert05.06.17, 15:11No: either you pass all perimeter of the square in 40 seconds, or only its side

Dolly03.06.17, 14:34Yes but statement is same so condition should also b same. Which is the correct understanding according to the situation? 10m in 40s or 40m in 40s

Assignment Expert02.06.17, 19:04It depends how to understand the condition: either 10 m in 40 sec, or 10*4 m in 40 sec.

Dolly02.06.17, 11:18Some of them are giving solution as considering the whole round of square field in 40s and calculating the answer as 14.14 m. I'm confused.

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